Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would effuse out of the same pinhole in 1.00 hour? Assume both gases are at the same temperature and pressure.

0.0019mol

Explanation:

To solve this problem, first we calculate for the rates of the effusion:

For CO2,

Number of mole = 0.00150 mol

Time = 1h

R1 = mol / time

R1 = 0.0015/1

R1 = 0.0015mol/h

Molar Mass of CO2 (M1) = 44.0 g/mol

For N2,

R2 = ?

Molar Mass of N2 (M2) = 28.0 g/mol

Using Graham's law, we can obtain Rate (R2) of N2 as follows:

R1/R2 = √ (M2/M1)

0.0015/R2 = √ (28/44)

0.0015/R2 = 0.7977

R2 = 0.0015 / 0.7977

R2 = 0.0019mol/h

But rate = mole / time

Time = 1h

Rate od N2 = 0.0019mol/h

0.0019 = mole / 1

Mole = 0.0019mol

Therefore, the number of mole of N2 is 0.0019mol